Question

在△ABC中，角A,B,C的对边依次为a，b，c，且A,B,C依次成等差数列。

（1）若AB向量*BC向量=-3/2,且b=√3，求a+c的值
（2)若A＜C，求2sin^2A+sin^2C的取值范围

Answer 1

(1) A,B,C依次成等差数列 -> 2B = A + C -> A + B + C = 3B = 180° -> B = 60°
AB向量*BC向量 = a * c * cos(180°- B) = -ac/2 = -3/2 -> ac = 3
cosB = (a² + c² - b²) / 2ac = (a² + c² - 3) / 6 = 1/2 -> a² + c² = 6
a + c = √(a + c)² = √(a² + 2ac + c²) = √12 = 2√3
(2) 设A = 60°- θ，B = 60°+ θ。0 ≤θ< 60°
2sin²A + sin²C
= 2sin²(60°- θ) + sin²(60°+ θ)
= 2(√3/2 * cosθ - 1/2 * sinθ)² + sin(√3/2 * cosθ + 1/2 * sinθ)²
= 9/4 * cos²θ - √3/2 * cosθ* sinθ + 3/4 * sin²θ
= 3/4 + 3/2 * cos²θ - √3/2 * cosθ* sinθ
= 3/4 + 3/2 * (1 + cos2θ) / 2 - √3/4 * sin2θ
= 3/2 + 3/4 * cos2θ - √3/4 * sin2θ
= 3/2 + √3/2 * (√3/2 * cos2θ - 1/2 * sin2θ)
= 3/2 + √3/2 *sin(60°- 2θ)
∵ 0 ≤θ< 60°
∴ -60°< 60°- 2θ≤ 60°
∴ -√3/2 < sin(60°- 2θ) ≤√3/2
∴ -3/4 < √3/2 *sin(60°- 2θ) ≤ 3/4
∴ 3/4 < 2sin²A + sin²C ≤ 9/4